$2y^2-x^2+x^3y=2$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{4y+x^3}{2x-3x^2y}$ (Choice B) B $\dfrac{2x-3x^2y}{4y+x^3}$ (Choice C) C $\dfrac{2x}{4y+3x^2}$ (Choice D) D $\dfrac{2x-4y}{3x^2}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $2y^2-x^2+x^3y=2$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} 2y^2-x^2+x^3y&=2 \\\\ \dfrac{d}{dx}(2y^2-x^2+x^3y)&=\dfrac{d}{dx}(2) \\\\ \dfrac{d}{dx}(2y^2)-\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(x^3y)&=0 \\\\ 4y\cdot\dfrac{dy}{dx}-2x+\left(3x^2\cdot y+x^3\cdot\dfrac{dy}{dx}\right)&=0 \\\\ 4y\cdot\dfrac{dy}{dx}-2x+3x^2y+x^3\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 4y\cdot\dfrac{dy}{dx}-2x+3x^2y+x^3\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(4y+x^3)&=2x-3x^2y \\\\ \dfrac{dy}{dx}&=\dfrac{2x-3x^2y}{4y+x^3} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{2x-3x^2y}{4y+x^3}$.